LeetCode #2486 — MEDIUM

Append Characters to String to Make Subsequence

Move from brute-force thinking to an efficient approach using two pointers strategy.

The Problem

Problem Statement

You are given two strings s and t consisting of only lowercase English letters.

Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Example 1:

Input: s = "coaching", t = "coding"
Output: 4
Explanation: Append the characters "ding" to the end of s so that s = "coachingding".
Now, t is a subsequence of s ("coachingding").
It can be shown that appending any 3 characters to the end of s will never make t a subsequence.

Example 2:

Input: s = "abcde", t = "a"
Output: 0
Explanation: t is already a subsequence of s ("abcde").

Example 3:

Input: s = "z", t = "abcde"
Output: 5
Explanation: Append the characters "abcde" to the end of s so that s = "zabcde".
Now, t is a subsequence of s ("zabcde").
It can be shown that appending any 4 characters to the end of s will never make t a subsequence.

Constraints:

1 <= s.length, t.length <= 10⁵
s and t consist only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given two strings s and t consisting of only lowercase English letters. Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Two Pointers · Greedy

Example 1

"coaching"
"coding"

Example 2

"abcde"
"a"

Example 3

"z"
"abcde"

Core Insight

What unlocks the optimal approach

Find the longest prefix of t that is a subsequence of s.
Use two variables to keep track of your location in s and t. If the characters match, increment both variables. Otherwise, only increment the variable for s.
The remaining characters in t must be appended to the end of s.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2486: Append Characters to String to Make Subsequence
class Solution {
    public int appendCharacters(String s, String t) {
        int n = t.length(), j = 0;
        for (int i = 0; i < s.length() && j < n; ++i) {
            if (s.charAt(i) == t.charAt(j)) {
                ++j;
            }
        }
        return n - j;
    }
}

// Accepted solution for LeetCode #2486: Append Characters to String to Make Subsequence
func appendCharacters(s string, t string) int {
	n, j := len(t), 0
	for _, c := range s {
		if j < n && byte(c) == t[j] {
			j++
		}
	}
	return n - j
}

# Accepted solution for LeetCode #2486: Append Characters to String to Make Subsequence
class Solution:
    def appendCharacters(self, s: str, t: str) -> int:
        n, j = len(t), 0
        for c in s:
            if j < n and c == t[j]:
                j += 1
        return n - j

// Accepted solution for LeetCode #2486: Append Characters to String to Make Subsequence
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2486: Append Characters to String to Make Subsequence
// class Solution {
//     public int appendCharacters(String s, String t) {
//         int n = t.length(), j = 0;
//         for (int i = 0; i < s.length() && j < n; ++i) {
//             if (s.charAt(i) == t.charAt(j)) {
//                 ++j;
//             }
//         }
//         return n - j;
//     }
// }

// Accepted solution for LeetCode #2486: Append Characters to String to Make Subsequence
function appendCharacters(s: string, t: string): number {
    let j = 0;
    for (const c of s) {
        if (c === t[j]) {
            ++j;
        }
    }
    return t.length - j;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.